# If $A\in \sigma(A_1,\cdots, A_n)$ and $A$ is independent of $A_1$, prove $A\in \sigma(A_2,\cdots, A_n)$.

I am not quite sure the whether following statement is correct:

If $$A\in \sigma(A_1,\cdots, A_n)$$ and $$A$$ is independent of $$A_1$$, then $$A\in \sigma(A_2,\cdots, A_n)$$. I can see that the sets in $$\sigma(A_1,\cdots, A_n)$$ which are independent of $$A$$ form a $$\sigma$$ algebra. But I just cannot prove it is indeed $$\sigma(A_2,\cdots, A_n)$$.

It looks quite reasonable, since intuitively $$A$$ is in $$\sigma(A_2,\cdots, A_n)$$ means $$A$$ can only be affected by $$A_2,\cdots, A_n$$. So I guess this should be true.

• Independence requires a probabilty measure. But the conclusion has nothing to do with measures. So the question hardly makes sense. Commented Oct 8, 2020 at 9:35
• @KaviRamaMurthy Thanks for comment. I see, this makes sense. Commented Oct 8, 2020 at 9:55

Let $$P(A)=0$$ but $$A \neq \emptyset$$. Let $$n=2, A_2=\emptyset$$ and $$A=A_1$$. This is a counter-example.